\(a,=\dfrac{12}{28}+\dfrac{22}{28}+\dfrac{19}{28}\\=\dfrac{12+22+19}{28}=\dfrac{53}{28}\\ b,=\dfrac{36}{45}+\dfrac{60}{45}+\dfrac{5}{45}\\ =\dfrac{36+60+5}{45}=\dfrac{101}{45}\\ c,=\dfrac{35}{70}+\dfrac{10}{70}+\dfrac{-14}{70}\\ =\dfrac{35+10+\left(-14\right)}{70}=\dfrac{31}{70}\\ d,=\dfrac{64}{96}+\dfrac{36}{96}+\dfrac{-40}{96}\\ =\dfrac{64+36+\left(-40\right)}{96}=\dfrac{60}{96}\\ e,=\dfrac{14}{16}+\dfrac{5}{16}+\dfrac{-12}{16}\\ =\dfrac{14+5+\left(-12\right)}{16}=\dfrac{7}{16}\)
a) Ta có: BCNN(7,14,28) = 28, nên:
\(\dfrac{3}{7}=\dfrac{3\times4}{7\times4}=\dfrac{12}{28}\) ; \(\dfrac{11}{14}=\dfrac{11\times2}{14\times2}=\dfrac{22}{28}\) ; \(\dfrac{19}{28}\)
⇒ \(\dfrac{3}{7}+\dfrac{11}{14}+\dfrac{19}{28}=\dfrac{12}{28}+\dfrac{22}{28}+\dfrac{19}{28}=\dfrac{53}{28}\)
b) Ta có: BCNN(5,3,9) = 45, nên:
\(\dfrac{4}{5}=\dfrac{4\times9}{5\times9}=\dfrac{36}{45}\) ; \(\dfrac{2}{3}=\dfrac{2\times15}{3\times15}=\dfrac{30}{45}\) ; \(\dfrac{1}{9}=\dfrac{1\times5}{9\times5}=\dfrac{5}{45}\)
⇒ \(\dfrac{4}{5}+\dfrac{2}{3}+\dfrac{1}{9}=\dfrac{36}{45}+\dfrac{30}{45}+\dfrac{5}{45}=\dfrac{71}{45}\)
a,\(\dfrac{3}{7}+\dfrac{11}{14}+\dfrac{19}{28}\)
\(=\dfrac{12}{28}+\dfrac{22}{28}+\dfrac{19}{28}\)
\(=\dfrac{12+22+19}{18}\)
\(=\dfrac{53}{28}\)
\(f,=\dfrac{3}{8}+\dfrac{4}{8}+\dfrac{8}{8}\\ =\dfrac{3+4+8}{8}=\dfrac{15}{8}\)
mik bấm máy nên chỉ có đáp án thôi nha
a 53/26
b 71/45
c 31/70
d 5/8
e 7/16
f 15/8
